Let N,X1,Y1,X2,Y2,… be independent random variables. The random variable N takes positive integer values and has mean a and variance r. The random variables Xi are independent and identically distributed with mean b and variance s, and the random variables Yi are independent and identically distributed with mean c and variance t. Let
A=∑i=1NXiandB=∑i=1NYi.
Find cov(A,B). Express your answer in terms of the given means and variances using standard notation .
cov(A,B)=
- unanswered
Find var(A+B). Express your answer in terms of the given means and variances using standard notation .
var(A+B)=
To find cov(A, B), we can start by calculating the covariance between the individual Xi and Yi random variables. Since Xi and Yi are independent and identically distributed, their covariance will be zero.
Now, let's consider the random variable Ci = Xi + Yi. The mean of Ci can be calculated as the sum of the means of Xi and Yi: b + c.
The variance of Ci can be calculated as the sum of the variances of Xi and Yi: s + t.
Since N is independent of Xi and Yi, we can use the properties of covariance to calculate cov(A, B) as follows:
cov(A, B) = cov(∑i=1NXi, ∑i=1NYi)
Using the properties of covariance, we know that cov(∑i=1NXi, ∑i=1NYi) = ∑i=1N cov(Xi, Yi)
Since Xi and Yi are independent and have zero covariance, cov(Xi, Yi) = 0 for all i.
Therefore, cov(A, B) = ∑i=1N 0 = 0.
So, cov(A, B) = 0.
To find var(A + B), we can use the properties of variance:
var(A + B) = var(∑i=1NXi + ∑i=1NYi)
Since the Xi and Yi random variables are independent, we can use the properties of variance to calculate var(A + B) as follows:
var(A + B) = var(∑i=1NXi) + var(∑i=1NYi)
Using the properties of variance, we know that var(∑i=1NXi) = N * var(Xi) and var(∑i=1NYi) = N * var(Yi).
Substituting the given variances for Xi and Yi, we have:
var(A + B) = N * s + N * t
Since N has variance r, we can substitute it into the equation:
var(A + B) = r * s + r * t
So, var(A + B) = r * (s + t).
Therefore, var(A + B) = r * (s + t).